Asked by Abbey
The vapor pressure of liquid chloroform, CHCl3, is 400.0 torr at 24.1°C and 100.0 torr at -6.3°C. What is ΔHvap (in the unit of kJ) of chloroform?
Answers
Answered by
bobpursley
ln(pressure)=ΔHvap/RT + C
so you have two conditions:
ln(400)= ΔHvap/(R*297)+C
ln(100)=ΔHvap/(R*266.7) + C
Hvap=(R*297)(ln400-C)
Hvap=(R*266.7)(ln100-C)
two equations, two unknowns. set the two equations equal, solve for C first, then go back into either equation, and solve for Hvap.
R is (8.314 J/mole K)
most solve this on graph paper, plotting ln(Pressure) vs 1/Temp, then measuring the slope, it will be Hvap, on the graph, C does not come into play.
so you have two conditions:
ln(400)= ΔHvap/(R*297)+C
ln(100)=ΔHvap/(R*266.7) + C
Hvap=(R*297)(ln400-C)
Hvap=(R*266.7)(ln100-C)
two equations, two unknowns. set the two equations equal, solve for C first, then go back into either equation, and solve for Hvap.
R is (8.314 J/mole K)
most solve this on graph paper, plotting ln(Pressure) vs 1/Temp, then measuring the slope, it will be Hvap, on the graph, C does not come into play.
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